rjacobson99 wrote:
EMPOWERgmatRichC wrote:
rjacobson99 wrote:
Can someone explain why / how you would know not to simplify the original provided equation (-20(t-5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:
-20(t-5)^2 + 500
-20(t^2 - 10t + 25) + 500 <-- foil
-20t^2 + 200t - 500 + 500 <-- distributed the -20
-20t^2 + 200t
20t(t + 10t) <-- simplified formula
Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past.
Hi rjacobson99,
In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with:
-20t^2 + 200t
(20t)(-t + 10)
(20t)(10 - t)
The maximum result will occur when t = 5.
GMAT assassins aren't born, they're made,
Rich
Hi Rich,
In the correct version of the formula (20t)(10-t), is there a way to determine the inflection point of t = 5 without having to plug in values from t = 1 to 6? I.e. is there a shortcut?
RJ
Hi rjacobson99,
Unfortunately, manipulating the equation in the way that you did places the variable "t" in both pairs of parentheses, so there isn't an 'obvious' solution to maximize the value. However, in the original equation, there IS an obvious Number Property that you can use...
N(t) = -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10.
In the given equation, notice how you have two "parts": the -20(something) and a +500. Here, to MAXIMIZE the value of N(t), we have to minimize the "impact" that the negative term - the -20(something) - has on the +500. By making that first part equal 0, we'll be left with 0 + 500. Mathematically, we have to make whatever is inside the parentheses equal 0....
(T-5) = 0
T = 5
GMAT assassins aren't born, they're made,
Rich
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